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Calculation 

f(x + y) = f(x) f′(y) + f′(x) f(x)

Put = x = y = 0

f(0) = f(0)f′(0) + f′(0)f(0) 

f′(0) = $\frac{1}{2}$

Put y = 0

f(x) = f(x) f′(0) + f′(x)f(0)

$f(x)=\frac{1}{2} f(x)+f^{\prime}(x)$

$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{f}(\mathrm{x})}{2}$

$\frac{d y}{d x}=\frac{y}{2} \Rightarrow \int \frac{d y}{y}=\int \frac{d x}{2}$

⇒ $\ell n y=\frac{x}{2}+c$

∵ f (0) = 1 ⇒ C = 0 

$\ell \mathrm{ny}=\frac{\pi}{2} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x} / 2}$

$\ell n \ \mathrm{f}(\mathrm{n})=\frac{\mathrm{n}}{2}$

$\sum_{\mathrm{n}=1}^{100} \ell \mathrm{f}(\mathrm{n})=\frac{1}{2} \sum_{\mathrm{n}=1}^{100} \mathrm{n}=\frac{5050}{2}$

= 2525

Hence option 2 is correct

Concept:

Counting reduced proper fractions:

• Set A: A = {1, 2, 3, …, 10}.
• Set B: B = {m/n : m,n ∈ A, m < n, gcd(m,n)=1}.
• Key idea: For a fixed denominator n, the number of integers
• 1 ≤ m < n with gcd(m,n)=1 equals φ(n).
• Important terms: Euler's totient function φ(n),
• reduced fraction, distinct fractions.
• Therefore: n(B) = Σ_n=2¹⁰ φ(n).

 

Calculation:

A = {1, 2, ….10}

\mathrm{B}\left{\frac{\mathrm{~m}}{\mathrm{n}}=\mathrm{m}, \mathrm{n} \in \mathrm{~A}, \mathrm{~m}<\mathrm{n}, \operatorname{gcd}(\mathrm{~m}, \mathrm{n})=1\right}

n(B)

\mathrm{n}=2 \quad\left{\frac{1}{2}\right}

⇒ φ(2) = 1

\mathrm{n}=3 \quad\left{\frac{1}{3}, \frac{2}{3}\right}

⇒ φ(3) = 2

\mathrm{n}=4 \quad\left{\frac{1}{4}, \frac{3}{4}\right}

⇒ φ(4) = 2

\mathrm{n}=5 \quad\left{\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\right}

⇒ φ(5) = 4

\mathrm{n}=6 \quad\left{\frac{1}{6}, \frac{5}{6}\right}

⇒ φ(6) = 2

\mathrm{n}=7 \quad\left{\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\right}

⇒ φ(7) = 6

\mathrm{n}=8 \quad\left{\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}\right}

⇒ φ(8) = 4

\mathrm{n}=9 \quad\left{\frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9}\right}  

⇒ φ(9) = 6 

\mathrm{n}=10 \quad\left{\frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10}\right}

⇒ φ(10) = 4

Hence n(B) = 31

∴ n(B) = 31.

Calculation

$y^{2}=2 \sqrt{3} x$

$(x-2 \sqrt{3})^{2}+y^{2}=(2 \sqrt{3})^{2}$

$A=\frac{π r^{2}}{2}-2 \int_{0}^{2 \sqrt{3}} \sqrt{2 \sqrt{3} x} d x$

$\frac{π(12)}{2}-2 \sqrt{2 \sqrt{3}} \frac{\left(\mathrm{x}^{3 / 2}\right)_{0}^{2 / 3}}{3 / 2}$

= 6π – 16

Hence option 3 is correct

Calculation

$P=\frac{\frac{6}{10} \times \frac{5}{9}}{\frac{4}{10} \times \frac{6}{9}+\frac{6}{10} \times \frac{5}{9}}=\frac{5}{9}$

m = 5, n = 9

m + n = 14

Calculation

 

be = 13, b = 5

a² = b² (e² – 1)

= b² e² – b²

= 169 – 25 = 144

$\ell(\mathrm{LR})=\frac{2 \mathrm{a}^{2}}{\mathrm{~b}}=\frac{2 \times 144}{5}=\frac{288}{5}$

Hence option 3 is correct

Ans. (34)

Sol.

f(x) is continuous and differentiable

at x = 1; LHL = RHL, LHD = RHD

–3a – 2 = a² + b, –6a = b

a = 2, 1; b = –12 

f(x)=\left{\begin{array}{ccc} -6 x^{2}-2 & ; & x<1 \ 4-12 x & ; & x \geq 1 \end{array}\right.

Area = $\int_{-\sqrt{3}}^{1}\left(-6 x^{2}-2+20\right) d x+\int_{1}^{2}(4-12 x+20) d x$

$16+12 \sqrt{3}+6=22+12 \sqrt{3}$

Calculation

$\int_{0}^{1}(1+x)^{11} d x=\left[C_{0} x+\frac{C_{1} x^{2}}{2}+\frac{C_{2} x^{3}}{3}+\ldots\right]_{0}^{1}$

$\frac{2^{12}-1}{12}=C_{0}+\frac{C_{1}}{2}+\frac{C_{2}}{3}+\frac{C_{3}}{4}+\ldots$

$\int_{-1}^{0}(1+x)^{11} d x=\left[C_{0} x+\frac{C_{1} x^{2}}{2}+\frac{C_{2} x^{3}}{3}+\ldots\right]_{-1}^{0}$

$\frac{1}{12}=C_{0}-\frac{C_{1}}{2}+\frac{C_{2}}{3}-\frac{C_{3}}{4}+\ldots$

$\frac{2^{12}-2}{12}=2\left(\frac{C_{1}}{2}+\frac{C_{3}}{4}+\frac{C_{5}}{6}+\ldots\right)$

\(\frac{\mathrm{C}{1}}{2}+\frac{\mathrm{C}{3}}{4}-\frac{\mathrm{C}_{5}}{6}+\ldots=\frac{2^{11}-1}{12}=\frac{2047}{12}\)

m - n = 2047 - 12 = 2035

Calculation

|A| = –2

det(3adj(–6adj(3A)))

= 33 det(adj(–adj(3A)))

= 3³ (–6)6 (det(3A))⁴

= 3²¹ × 2¹⁰

m + n = 10

mn = 21

m = 7 ; n = 3

4m + 2n = 28 + 6 = 34

Calculation

Point B

(3λ + 1, –λ + 1, –1) ≡ (2µ + 2, 0, αµ – 4)

3λ + 1 = 2µ + 2

–λ + 1 = 0

–1 = αµ – 4

λ = 1, µ = 1, α = 3

B(4, 0, –1)

Let Point ‘P’ is (2δ + 2, 0, 3δ – 4)

Dr’s of AP < 2δ + 1, –1, 3δ – 3 >

$\mathrm{AP} \perp \mathrm{~L}_{2} \Rightarrow δ=\frac{7}{13}$

$\mathrm{P}\left(\frac{40}{13}, 0, \frac{-31}{13}\right)$

$2 \sigma δ(\mathrm{~PB})^{2}=26 \times 3 \times\left(\frac{144}{169}+\frac{324}{169}\right)$

= 216

Calculation

$\overrightarrow{\mathrm{c}}=\left(\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}\right) \frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}$

⇒ $\left(\frac{\lambda+8}{9}\right)(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$

$|\vec{a}+\vec{c}|-=7 \Rightarrow \lambda=4$

Area of parallelogram

= $|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}|=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ \frac{4}{3} & \frac{8}{3} & \frac{8}{3} \ 4 & 0 & 4 \end{array}\right|$

= 16